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Let $f \in C^k(0,1)$ and assume that the $k$-th derivative is $\alpha$-Hölder continuous. Assume that $f(x) = 0$ in a fixed interval $(a,b) \subset (0,1)$. Can we characterize (or at least find some examples of) non-constant functions $f$ as above such that $$|f^{(k)}|_{C^{0,\alpha}(0,1)} \le \Vert f \Vert_{L^1(0,1)},$$ where $|g|_{C^{0,\alpha}} = \sup_{x,y \in (0,1)} \frac{|g(x)-g(y)|}{|x-y|^\alpha}$?

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    $\begingroup$ Take any smooth $f$ and add a sufficiently large constant to it. Or am I missing something? $\endgroup$ Jan 24 at 23:00
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    $\begingroup$ @MateuszKwaśnicki Sorry, I forgot to add the assumption that $f(x) = 0$ ona given interval $(a,b) \subset (0,1)$ (so we cannot just add a constant to make its $L^1$ norm larger without also making the $|\cdot|_{C^{0,\alpha}}$ seminorm larger) $\endgroup$
    – Hiro
    Jan 24 at 23:07
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The answer is no.

Indeed, let $t:=\alpha\in(0,1)$ and $c:=\|f\|_1:=\|f\|_{L^1(0,1)}\in(0,\infty)$. Suppose that $f\in C^k(0,1)$ and $f=0$ on $(a,b)$, with $0\le a<b\le1$. Suppose that the inequality in question holds.

Then for $x\in[b,1]$ we have $|f^{(k)}(x)|=|f^{(k)}(x)-f^{(k)}(b)|\le c(x-b)^t\le c$, $|f^{(k-1)}(x)|\le c(x-b)\le c$, $\ldots$, $|f(x)|\le c$, $\int_b^1|f|\le c(1-b)$.

Similarly, $\int_0^a|f|\le ca$. So, $$c=\|f\|_1=\int_0^a|f|+\int_b^1|f|\le ca+c(1-b)<c,$$ a contradiction. $\Box$

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